# Question #6fd67

Nov 1, 2015

(B) KInetic energy at P is equal to potential energy at Q

#### Explanation:

The idea here is that the energy that the ball has in point $P$ will be converted to another type of energy as it reaches point $Q$.

So, the ball starts at a point $P$. Before being thrown upward, it will only have potential energy associated with the height at which point $P$ is located. Let's say that point $P$ is located at $p$ meters above the ground. You can thus say that

${E}_{P} = m \cdot g \cdot p \to$ potential energy at p meteres above the ground

Now, what would the ball need in order to go from a point $P$ to a higher point $Q$?

SInce it's being thrown, an initial velocity directed straight upward must be applied to the ball. This means that at point $P$, the energy of the ball must contain a kinetic component as well.

Let's assume that the ball is being thrown with a velocity $v$. Its energy at point $P$ will be

${E}_{P} = m \cdot g \cdot p + \frac{1}{2} \cdot m {v}^{2}$

What will happen as the ball moves upward?

Since it's being acted upon by gravity, which always points downward, its initial velocity will decrease as the height of the ball increases.

This means that its kinetic energy will decrease as well. At its highest point, the ball's velocity will be equal to zero.

SInce the height at which the ball is located increases, its potential energy will increase as well. Let's assume that point $Q$ is located at $q$ meters above the ground.

In point $Q$, the energy of the ball will be

${E}_{Q} = m \cdot g \cdot q + \frac{1}{2} \cdot m \cdot {0}^{2} = m \cdot g \cdot q$

Since the ball started from $p$ meters and reached $q$ meters, it follows that it travelled $q - p$ meters upwards.

Since the energy of the ball must be conserved in points $P$ and $Q$, you can say that

${E}_{P} = {E}_{Q}$

$m \cdot g \cdot p + \frac{1}{2} \cdot m \cdot {v}^{2} = m \cdot g \cdot q$

If you rearrange this equation, you'll get

$\frac{1}{2} \cdot m \cdot {v}^{2} = m \cdot g \cdot q - m \cdot g \cdot p$

$\frac{1}{2} \cdot m \cdot {v}^{2} = m \cdot g \cdot \left(q - p\right)$

This tells you that the kinetic energy that the ball had in point $P$ was converted to potential energy in point $Q$.

If you take point $P$ to be ground level, then you can see that

$\frac{1}{2} \cdot m \cdot {v}^{2} = m \cdot g \cdot q \to$ kinetic energy at p is equal to potential energy at q